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We all know the elementary school definition of the trigonometric functions: In a right triangle, $\cos$ is defined to be the ratio of the adjacent and the hypotenuse, etc.

But, I observed that this definition has one "problem": For a given angle, there are infinitely many right triangles having that angle. How does one prove that the ratio $$\frac{\text{adjacent}}{\text{hypotenuse}}$$ is independent of the right triangle used? The same goes for the other ratios.

Proving the "opposite direction" seems easy: the angle between vectors $u$ and $v$ is $$\arccos \frac{u\cdot v}{|u||v|}$$ and if the vectors $u'$ and $v'$ are constant multiples of $u$ and $v$, respectively, then we see that the common factor cancels out nicely in the $\arccos$ formula.

Nomas2
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    "How does one prove that the ratio [...] is independent of the right triangle used?" By showing that triangles with equal corresp angles have proportional corresp sides, and vice-versa. As this answer notes, Euclid proved the equivalence in Prop 4 and Prop 5 of Book 6 of Elements. Also, rt triangles sharing an acute angle have all corresp angles equal, as any triangle's angle-sum is $180^\circ$. – Blue Nov 05 '23 at 19:43
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    Importantly, this kind of thing only works in Euclidean geometry. In non-Euclidean (ie, "hyperbolic" and "spherical") geometries, there is no universal angle-sum; thus, right triangles sharing an acute angle need not share their other acute angle. Moreover, in these geometries, AAA joins SSS, SAS, and ASA as a congruence pattern, so if both acute angles match, then those right triangles are identical. As a result, ratios in right triangles do not —can not— form the basis of non-Euclidean trig. – Blue Nov 05 '23 at 19:54
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    In three words: the intercept theorem. It's an important theorem in Euclidean geometry describing the conditions of similarity (which is the property responsible for the independence of trig ratios to the sizes of the triangles used) – H. sapiens rex Nov 05 '23 at 22:11

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